1. How to solve?

2/1 - 1/a+2 = a+18/4a^2-16

LCD: 4(a-2)(a+2)

2. Re: How to solve?

Originally Posted by benjixbunn
2/1 - 1/a+2 = a+18/4a^2-16

LCD: 4(a-2)(a+2)

I have a feeling from the 2/1 in the beginning that there must be a mistake. Please recheck this and rewrite the equation using brackets where they are needed.

3. Re: How to solve?

I meant 2 - 1/a+2 = a+18/4a^2-16

4. Re: How to solve?

Originally Posted by benjixbunn
I meant 2 - 1/a+2 = a+18/4a^2-16
Assuming you meant \displaystyle \displaystyle \begin{align*} 2 - \frac{1}{a + 2} = \frac{a + 18}{4a^2 - 16} \end{align*}, which it could very well not be because you didn't use brackets like you were asked to... First note that \displaystyle \displaystyle \begin{align*} a \neq \pm 2 \end{align*} (why?), then

\displaystyle \displaystyle \begin{align*} 2 - \frac{1}{a + 2} &= \frac{a + 18}{4a^2 - 16} \\ 2 - \frac{1}{a + 2} &= \frac{a + 18}{4(a - 2)(a + 2)} \\ \frac{2\cdot 4(a - 2)(a + 2)}{4(a - 2)(a + 2)} - \frac{4(a - 2)}{4(a - 2)(a + 2)} &= \frac{a + 18}{4(a - 2)(a + 2)} \\ \frac{8(a - 2)(a + 2) - 4(a - 2)}{4(a - 2)(a + 2)} &= \frac{a + 18}{4(a - 2)(a + 2)} \\ 8(a - 2)(a + 2) - 4(a - 2) &= a + 18 \end{align*}

Can you continue?

5. Re: How to solve?

Originally Posted by Prove It
Assuming you meant \displaystyle \displaystyle \begin{align*} 2 - \frac{1}{a + 2} = \frac{a + 18}{4a^2 - 16} \end{align*}, which it could very well not be because you didn't use brackets like you were asked to... First note that \displaystyle \displaystyle \begin{align*} a \neq \pm 2 \end{align*} (why?), then

\displaystyle \displaystyle \begin{align*} 2 - \frac{1}{a + 2} &= \frac{a + 18}{4a^2 - 16} \\ 2 - \frac{1}{a + 2} &= \frac{a + 18}{4(a - 2)(a + 2)} \\ \frac{2\cdot 4(a - 2)(a + 2)}{4(a - 2)(a + 2)} - \frac{4(a - 2)}{4(a - 2)(a + 2)} &= \frac{a + 18}{4(a - 2)(a + 2)} \\ \frac{8(a - 2)(a + 2) - 4(a - 2)}{4(a - 2)(a + 2)} &= \frac{a + 18}{4(a - 2)(a + 2)} \\ 8(a - 2)(a + 2) - 4(a - 2) &= a + 18 \end{align*}

Can you continue?

Can you explain to me please??

6. Re: How to solve?

Originally Posted by moonwoo
Can you explain to me please??
Prove It found a common denominator for all the terms in the equation ... once that is accomplished, the numerator forms an equation that can be solved, keeping in mind the restrictions in values for $\displaystyle a$.